{"id":126308,"date":"2023-09-10T12:53:00","date_gmt":"2023-09-10T07:23:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126308"},"modified":"2023-11-28T11:04:44","modified_gmt":"2023-11-28T05:34:44","slug":"rd-sharma-class-9-solutions-chapter-8-exercise-8-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-8-exercise-8-1\/","title":{"rendered":"RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126318\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-8-Exercise-8.1.jpg\" alt=\"RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-8-Exercise-8.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-9-Solutions-Chapter-8-Exercise-8.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1: <\/strong>RD Sharma Solutions for Class 9 Maths are given here for Chapter 8 \u2013 Lines and Angles. In order to make concepts easier for students, we have provided RD Sharma Class 9 Maths Chapter 8 solutions to help them go through several solved exercises and, at the same time, practice the questions. RD Sharma Solutions for Class 9 can help students discover new ways to solve difficult problems.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"ez-toc-toggle-icon-1\"><label for=\"item-69d07260d1ee1\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/label><input  type=\"checkbox\" id=\"item-69d07260d1ee1\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-8-exercise-8-1\/#download-rd-sharma-class-9-solutions-chapter-8-exercise-81-pdf\" title=\"Download RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1 PDF\">Download RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1 PDF<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-8-exercise-8-1\/#access-answers-of-rd-sharma-for-class-9-solutions-chapter-8-exercise-81\" title=\"Access answers of RD Sharma for Class 9 Solutions Chapter 8 Exercise 8.1\">Access answers of RD Sharma for Class 9 Solutions Chapter 8 Exercise 8.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-8-exercise-8-1\/#faqs-on-rd-sharma-class-9-solutions-chapter-8-exercise-81\" title=\"FAQs on RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1\">FAQs on RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-8-exercise-8-1\/#from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-for-chapter-8-exercise-81\" title=\"From where can I download the PDF of RD Sharma Class 9 Solutions for Chapter 8 Exercise 8.1?\">From where can I download the PDF of RD Sharma Class 9 Solutions for Chapter 8 Exercise 8.1?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-8-exercise-8-1\/#how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-for-class-9-solutions-chapter-8-exercise-81\" title=\"How much does it cost to download the PDF of RD Sharma for Class 9 Solutions Chapter 8 Exercise 8.1?\">How much does it cost to download the PDF of RD Sharma for Class 9 Solutions Chapter 8 Exercise 8.1?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-8-exercise-8-1\/#can-i-access-the-rd-sharma-class-9-solutions-for-chapter-8-exercise-81-pdf-offline\" title=\"Can I access the RD Sharma Class 9 Solutions for Chapter 8 Exercise 8.1\u00a0PDF offline?\">Can I access the RD Sharma Class 9 Solutions for Chapter 8 Exercise 8.1\u00a0PDF offline?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-9-solutions-chapter-8-exercise-81-pdf\"><\/span><strong>Download RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1 PDF<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/8.1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/8.1.pdf\", \"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-for-class-9-solutions-chapter-8-exercise-81\"><\/span><strong>Access answers of RD Sharma for Class 9 Solutions Chapter 8 Exercise 8.1<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p><strong>Question 1: Write the complement of each of the following angles:<\/strong><\/p>\n<p><strong>(i)20<sup>0<\/sup><\/strong><\/p>\n<p><strong>(ii)35<sup>0<\/sup><\/strong><\/p>\n<p><strong>(iii)90<sup>0<\/sup><\/strong><\/p>\n<p><strong>(iv) 77<sup>0<\/sup><\/strong><\/p>\n<p><strong>(v)30<sup>0<\/sup><\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p><strong>(i)<\/strong>&nbsp;The sum of an angle and its complement = 90<strong><sup>0<\/sup><\/strong><\/p>\n<p>Therefore, the complement of 20<strong><sup>0&nbsp;<\/sup><\/strong>= 90<strong><sup>0<\/sup><\/strong>&nbsp;\u2013 20<strong><sup>0<\/sup><\/strong>&nbsp;= 70<strong><sup>0<\/sup><\/strong><\/p>\n<p><strong>(ii)<\/strong>&nbsp;The sum of an angle and its complement = 90<strong><sup>0<\/sup><\/strong><\/p>\n<p>Therefore, the complement of 35\u00b0 = 90\u00b0 \u2013 35\u00b0 = 55<\/p>\n<p><strong>(iii)<\/strong>&nbsp;The sum of an angle and its complement = 90<strong><sup>0<\/sup><\/strong><\/p>\n<p>Therefore, the complement of 90<strong><sup>0<\/sup><\/strong>&nbsp;= 90<strong><sup>0<\/sup><\/strong>&nbsp;\u2013 90<strong><sup>0<\/sup><\/strong>&nbsp;= 0<strong><sup>0<\/sup><\/strong><\/p>\n<p><strong>(iv)<\/strong>&nbsp;The sum of an angle and its complement = 90<strong><sup>0<\/sup><\/strong><\/p>\n<p>Therefore, the complement of 77<strong><sup>0<\/sup><\/strong>&nbsp;= 90\u00b0 \u2013 77<strong><sup>0<\/sup><\/strong>&nbsp;= 13<strong><sup>0<\/sup><\/strong><\/p>\n<p><strong>(v)<\/strong>&nbsp;The sum of an angle and its complement = 90<strong><sup>0<\/sup><\/strong><\/p>\n<p>Therefore, the complement of 30<strong><sup>0<\/sup><\/strong>&nbsp;= 90<strong><sup>0<\/sup><\/strong>&nbsp;\u2013 30<strong><sup>0<\/sup><\/strong>&nbsp;= 60<strong><sup>0<\/sup><\/strong><\/p>\n<p><strong>Question 2: Write the supplement of each of the following angles:<\/strong><\/p>\n<p><strong>(i) 54<sup>0<\/sup><\/strong><\/p>\n<p><strong>(ii) 132<sup>0<\/sup><\/strong><\/p>\n<p><strong>(iii) 138<sup>0<\/sup><\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p><strong>(i)<\/strong>&nbsp;The sum of an angle and its supplement = 180<strong><sup>0<\/sup><\/strong>.<\/p>\n<p>Therefore supplement of angle 54<strong><sup>0<\/sup><\/strong>&nbsp;= 180<strong><sup>0<\/sup><\/strong>&nbsp;\u2013 54<strong><sup>0<\/sup><\/strong>&nbsp;= 126<strong><sup>0<\/sup><\/strong><\/p>\n<p><strong>(ii)<\/strong>&nbsp;The sum of an angle and its supplement = 180<strong><sup>0<\/sup><\/strong>.<\/p>\n<p>Therefore supplement of angle 132<strong><sup>0<\/sup><\/strong>&nbsp;= 180<strong><sup>0<\/sup><\/strong>&nbsp;\u2013 132<strong><sup>0<\/sup><\/strong>&nbsp;= 48<strong><sup>0<\/sup><\/strong><\/p>\n<p><strong>(iii)<\/strong>&nbsp;The sum of an angle and its supplement = 180<strong><sup>0<\/sup><\/strong>.<\/p>\n<p>Therefore supplement of angle 138<strong><sup>0&nbsp;<\/sup><\/strong>= 180<strong><sup>0&nbsp;<\/sup><\/strong>\u2013 138<strong><sup>0<\/sup><\/strong>&nbsp;= 42<strong><sup>0<\/sup><\/strong><\/p>\n<p><strong>Question 3: If an angle is 28<sup>0<\/sup> less than its complement, find its measure.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the measure of any angle is \u2018 a \u2018 degrees<\/p>\n<p>Thus, its complement will be (90 \u2013 a)<strong><sup>&nbsp;0<\/sup><\/strong><\/p>\n<p>So, the required angle = Complement of a \u2013 28<\/p>\n<p>a = ( 90 \u2013 a ) \u2013 28<\/p>\n<p>2a = 62<\/p>\n<p>a = 31<\/p>\n<p>Hence, the angle measured is 31<strong><sup>0<\/sup><\/strong>.<\/p>\n<p><strong>Question 4: If an angle is 30\u00b0 more than one-half of its complement, find the measure of the angle.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Let an angle measured by \u2018 a \u2018 in degrees<\/p>\n<p>Thus, its complement will be (90 \u2013 a)<strong><sup>&nbsp;0<\/sup><\/strong><\/p>\n<p>Required Angle = 30<strong><sup>0<\/sup><\/strong>&nbsp;+ complement\/2<\/p>\n<p>a = 30<strong><sup>0<\/sup><\/strong>&nbsp;+ ( 90 \u2013 a )<strong><sup>&nbsp;0<\/sup><\/strong>&nbsp;\/ 2<\/p>\n<p>a + a\/2 = 30<strong><sup>0<\/sup><\/strong>&nbsp;+ 45<strong><sup>0<\/sup><\/strong><\/p>\n<p>3a\/2 = 75<strong><sup>0<\/sup><\/strong><\/p>\n<p>a = 50<strong><sup>0<\/sup><\/strong><\/p>\n<p>Therefore, the measure of the required angle is 50<strong><sup>0<\/sup><\/strong>.<\/p>\n<p><strong>Question 5: Two supplementary angles are in the ratio 4:5. Find the angles.<\/strong><\/p>\n<p><strong>Solution<\/strong>:<\/p>\n<p>Two supplementary angles are in the ratio 4:5.<\/p>\n<p>Let us say, the angles are 4a and 5a (in degrees)<\/p>\n<p>Since angles are supplementary angles;<\/p>\n<p>Which implies, 4a + 5a = 180<strong><sup>0<\/sup><\/strong><\/p>\n<p>9a = 180<strong><sup>0<\/sup><\/strong><\/p>\n<p>a = 20<strong><sup>0<\/sup><\/strong><\/p>\n<p>Therefore, 4a = 4 (20) = 80<strong><sup>0&nbsp;<\/sup><\/strong>and<\/p>\n<p>5(a) = 5 (20) = 100<strong><sup>0<\/sup><\/strong><\/p>\n<p>Hence, the required angles are 80\u00b0 and 100<strong><sup>0<\/sup><\/strong>.<\/p>\n<p><strong>Question 6: Two supplementary angles differ by 48<sup>0<\/sup>. Find the angles?<\/strong><\/p>\n<p><strong>Solution<\/strong>: Given: Two supplementary angles differ by 48<strong><sup>0<\/sup><\/strong>.<\/p>\n<p>Consider a<strong><sup>0<\/sup><\/strong> to be one angle then its supplementary angle will be equal to (180 \u2013 a)<strong><sup>&nbsp;0<\/sup><\/strong><\/p>\n<p>According to the question;<\/p>\n<p>(180 \u2013 a ) \u2013 x = 48<\/p>\n<p>(180 \u2013 48 ) = 2a<\/p>\n<p>132 = 2a<\/p>\n<p>132\/2 = a<\/p>\n<p>Or a = 66<strong><sup>0<\/sup><\/strong><\/p>\n<p>Therefore, 180 \u2013 a = 114<strong><sup>0<\/sup><\/strong><\/p>\n<p>Hence, the two angles are 66<strong><sup>0<\/sup><\/strong>&nbsp;and 114<strong><sup>0<\/sup><\/strong>.<\/p>\n<p><strong>Question 7: An angle is equal to 8 times its complement. Determine its measure.<\/strong><\/p>\n<p><strong>Solution:<\/strong>&nbsp;Given: Required angle = 8 times of its complement<\/p>\n<p>Consider a<strong><sup>0<\/sup><\/strong> to be one angle then its complementary angle will be equal to (90 \u2013 a)<strong><sup>&nbsp;0<\/sup><\/strong><\/p>\n<p>According to the question;<\/p>\n<p>a = 8 times its complement<\/p>\n<p>a = 8 ( 90 \u2013 a )<\/p>\n<p>a = 720 \u2013 8a<\/p>\n<p>a + 8a = 720<\/p>\n<p>9a = 720<\/p>\n<p>a = 80<\/p>\n<p>Therefore, the required angle is 80<strong><sup>0<\/sup><\/strong>.<\/p>\n<p>This is the complete blog on RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1. To know more about the <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a> Class 9 Maths exam, ask in the comments.&nbsp;<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-9-solutions-chapter-8-exercise-81\"><\/span><strong>FAQs on RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631212825766\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-class-9-solutions-for-chapter-8-exercise-81\"><\/span>From where can I download the PDF of RD Sharma Class 9 Solutions for Chapter 8 Exercise 8.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631212885392\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-for-class-9-solutions-chapter-8-exercise-81\"><\/span>How much does it cost to download the PDF of RD Sharma for Class 9 Solutions Chapter 8 Exercise 8.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631212966577\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-class-9-solutions-for-chapter-8-exercise-81-pdf-offline\"><\/span>Can I access the RD Sharma Class 9 Solutions for Chapter 8 Exercise 8.1\u00a0PDF offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline as well.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1: RD Sharma Solutions for Class 9 Maths are given here for Chapter 8 \u2013 Lines and Angles. In order to make concepts easier for students, we have provided RD Sharma Class 9 Maths Chapter 8 solutions to help them go through several solved exercises and, at &#8230; <a title=\"RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-9-solutions-chapter-8-exercise-8-1\/\" aria-label=\"More on RD Sharma Class 9 Solutions Chapter 8 Exercise 8.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":126318,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411],"tags":[3086,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126308"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=126308"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126308\/revisions"}],"predecessor-version":[{"id":513219,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126308\/revisions\/513219"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126318"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126308"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126308"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126308"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}